We can define POWER as the work that can be performed in a given
period of time. Thus, we can say that it is the speed with which a given
work is performed. Your unit of measure is watt or
joule / second.

Another way of expressing the unit of measure of power is to say that if one
joule is used to transfer one coulomb of load in one second,
through a device, then the energy transfer rate is equal to one watt.

As coulomb / second is defined as electric current
(which is measured in ampère), it becomes easier to express power as the product
between electric current and voltage. And we know from the
law of ohm, the relationship between electric voltage, electric current, and
resistance. Therefore, we can write the equations that allow us to calculate the power
in a circuit, or:

eq. 07-01

eq. 07-02

eq. 07-03

where the variables are:

P - power, whose unit of measure is watt

V - electrical voltage, whose unit of measure is volt

I - electric current, whose unit of measure is ampère

R - electrical resistance, whose unit of measure is ohm

2. Convention

In the study of power we must establish an adequate convention in such a way that
through the analysis of the sign of magnitude, establish whether the device
is receiving or supplying electrical power to the circuit.

Thus, we say that a device is receiving power, if it has signal POSITIVE
in its numerical value. For example, an electrical resistance is always receiving power.
Therefore, it is common to hear people saying that "resistance is
dissipating energy in the form of heat." As the electric resistance is receiving power,
and it does not have the capacity to store energy, then over time (E = P t)
releases this energy in the form of heat, also known as the joule effect.
The electric shower is a typical example of this property. Use this effect to heat water.

So when we read that such a device has, for example, 3 000 watts of power,
this means that it will absorb 3 000 watts of power from one
other device that should provide that power to it. In case,
the electric power company that supplies the residence. In case of electric shower, it absorbs
(dissipates) + 3 000 watts and the device providing this power
to the electric shower, we represent - 3 000 watts.

Notice that in any situation, in an electric circuit, the algebraic sum
of the received and dissipated powers will always equal ZERO.

3. Power Analysis in a Circuit

In the Figure 07-01, as an example, we present an electric circuit where we shows the values
of the electric current that circulates through its components, as well as the direction of the same. Pay attention
that the electric current leaves by the positive pole of the voltage source of 28 volts.
Then, by the adopted convention this voltage source is providing power to the circuit.
Therefore, we represent the numerical value of the power provided by a NEGATIVE number. The same happens with
voltage source of 20 volts, where the electric current also leaves by the
positive pole. This means that
this voltage source is supplying power to the circuit. Thus, we will represent
the numeric value of power delivered by this source as a NEGATIVE number. The electrical resistances that form the
circuit, obviously, because they are passive devices, receive power from the sources
dissipating it in the form of heat (joule effect). Therefore, in electrical resistors
the numeric value of power will always be POSITIVE.

To make it clear how we should proceed, we will make a power balance of the circuit.

Power Balance

First, let us calculate the dissipated powers in all resistors.

P_{4} = 4 ( i_{1} )^{2} = 4 6.5^{2} = 169 W

P_{2} = 2 ( i_{2} )^{2} = 2 1^{2} = 2 W

P_{4} = 4 ( i_{3} )^{2} = 4 5.5^{2} = 121 W

Summing up algebraically the dissipated powers by the resistors, we find:

P^{+} = 169 + 2 + 121 = 292 W

For the sources, notice that the current leaves by the positive pole of the 28 volts source,
that is, it is supplying power to the circuit, so its value is negative, or:

P_{28} = - 28 |i_{1}| = - 28 6.5 = - 182 W

For the voltage source of 20 volts, the current also leaves at
positive pole, supplying power to the circuit. Therefore, its value is also negative. So:

P_{20} = - 20 |i_{3}| = - 20 5.5 = - 110 W

Now, we can add algebraically the powers provided by the sources and we find:

P^{-} = - 182 - 110 = - 292 W

Finally, we know that the algebraic sum of the power supplied and dissipated
in a circuit must be equal to ZERO, or:

∑ P = P^{+} + P^{-} = 292 - 292 = 0 W

Thus, through a practical example, we show how to use the sign convention
to calculate the power involved in a circuit.

4. Maximum Power Transfer Theorem

One of the designer's concerns is to know under what conditions we can transfer the maximum power to a load
which is connected at the output of a circuit.
To solve this problem we have called the Maximum Power Transfer Theorem .
This theorem starts from the assumption that we have a circuit formed by an energy generator with a given
known internal resistance, which we will call R_{i}. When we put
a load R_{L} on its output terminals, we are interested in knowing the value of this load
so that there is maximum power transfer to it.

We can easily prove this theorem by starting from a resistive voltage divider circuit,
where we must consider one of the resistances to be the load and the other, the internal resistance
of the generator. For those interested in seeing proof
click here!

Let us anticipate that for this event to happen the following relation must be satisfied:

eq. 07-04

In other words: whenever the value of load is exactly equal to the value of
internal resistance of the generator, we have guaranteed the necessary condition for there to be
the maximum power transfer to the load.

At no point does this theorem assert that the inverse path is valid.
Do not make this "Mistake". That is, if we have a load of known value
the choice of the generator should not fall on the one having the internal resistance EQUAL
to load, as this will certainly not guarantee the maximum power transfer.
It is obvious that for this to happen, we must have the internal resistance of the generator equal to ZERO.
So, stay tuned to the details.

Remember: the maximum power transfer theorem is known as a theorem
ONE WAY. Do not transform it, on its own, into a theorem TWO WAY.

If there are still doubts regarding the above statement, let's take as an example the
problem 7-1 (if you are interested in seeing your solution click here!).
We will make a change in the statement.
Let's keep the value of R_{L} constant, however we will vary the value
of R_{i}.

Example statement: Calculate the power dissipated by the load R_{L} = 50 Ω
when R_{i} assumes the following
values: 0, 5, 10, 30, 50, 70 and 100 ohms. Calculate the value of R_{i}
so that R_{L} dissipate the maximum power. Consider the circuit
shown in the Figure 07-02.

Let us through this example, show how the power dissipated by the load varies
for different values of R_{i}. We will
table for better understanding. The steps for mounting the table are
determine the current I when R_{i} assumes the value
100 ohms and then calculate for subsequent values. So, first let's go
resolve to the value of 100 ohms and then repeat the same steps
for the other values of R_{i}, not forgetting that the source voltage,
for our example, is 50 volts.

I = 50 / R_{i} + R_{L} = 50 / (100 + 50) = 0.333 A

Now that we have the electric current flowing through R_{L} = 50 ω
just apply the equation that allows us to calculate the power, or:

P_{RL} = R_{L} I^{2}= 50 0.333^{2} = 5.56 W

Doing similarly for the other values of R_{i},
we can assemble the table as shown below.

Table 07-01

Valor de R_{i} (Ω)

Corrente I (A)

Potência em R_{L} (W)

100

0.33

5.56

70

0.42

8.68

50

0.50

12.50

30

0.625

19.53

10

0.83

34.72

5

0.91

41.32

0

1.00

50.00

In the Table 07-01 we can see that when R_{i} decreases,
the power dissipated in R_{L} grows.
See the Figure 07-03, the power dissipated graph in R_{L} when
we change the value of R_{i}. For the realization of this graph we use the eq. 7-1,
already shown above.
For this case, we use V = 50 volts and R_{L} = 50 ohms.

In the Figure 07-03 we see that the peak power dissipated by the load occurs exactly when
we have the value of R_{i} = 0. As we stated earlier. Therefore,
we prove that the maximum power transfer theorem is a
theorem ONE WAY, that is, it only holds when R_{i} is
CONSTANT and the load is VARIABLE (case of problem 7-1
See Here! ).
In this case we analyze (constant load and R_{i} variable),
NO applies.